9.2 Rotational Dynamics and Moment of Inertia

Rotational dynamics explains how and why objects speed up, slow down, or maintain their spin. Just as Newton's Second Law connects force, mass, and acceleration for straight-line motion, its rotational counterpart connects torque, moment of inertia, and angular acceleration. Mastering these parallels is the fastest way to get comfortable with rotation problems.

The central equation is $\sum \tau = I \alpha$. Once you can identify the torques acting on a system and calculate its moment of inertia, you can predict rotational acceleration the same way you'd predict linear acceleration with $\sum F = ma$.

Rotational Motion Fundamentals

Newton's Second Law in rotation

The rotational version of Newton's Second Law is:

$\sum \tau = I \alpha$

Each quantity maps directly onto its linear counterpart: torque $\tau$ replaces force $F$, moment of inertia $I$ replaces mass $m$, and angular acceleration $\alpha$ replaces linear acceleration $a$. If you're comfortable with $\sum F = ma$, the logic here is identical.

To apply it, you follow the same strategy as linear problems:

1. Identify every torque acting on the object (including sign/direction).
2. Determine the object's moment of inertia about the rotation axis.
3. Set $\sum \tau = I \alpha$ and solve for the unknown.

The rotational kinematics equations also mirror their linear versions. Use these when angular acceleration is constant:

• $\omega = \omega_0 + \alpha t$
• $\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2$
• $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$

These work exactly like $v = v_0 + at$ and the other linear kinematics equations, just with angular variables swapped in.

Torque and net torque

Torque is what causes angular acceleration, the rotational equivalent of force. It's calculated as:

$\tau = r F \sin \theta$

Here, $r$ is the distance from the rotation axis to where the force is applied, $F$ is the magnitude of the force, and $\theta$ is the angle between the force vector and the position vector (the line from the axis to the point of application). When the force is perpendicular to that line, $\sin \theta = 1$ and you get maximum torque. When the force points straight along that line, $\sin \theta = 0$ and there's no torque at all.

Torque is a vector quantity, measured in Newton-meters (N·m). Its direction is determined by the right-hand rule: curl your fingers in the direction the force would cause rotation, and your thumb points along the torque vector. For 2D problems, you'll typically just assign positive to counterclockwise and negative to clockwise (or vice versa, as long as you're consistent).

To find net torque, sum all individual torques acting on the object, keeping track of their signs. Only the net torque goes into $\sum \tau = I \alpha$.

Moment of Inertia and Rotational Dynamics

Moment of inertia concept

Moment of inertia ($I$) measures how much an object resists changes in its rotational motion. It plays the same role in rotation that mass plays in linear motion: a larger $I$ means the object is harder to spin up or slow down for a given torque.

What makes moment of inertia different from mass is that it depends not just on how much mass there is, but on where that mass is located relative to the rotation axis. Mass concentrated far from the axis contributes much more to $I$ than the same mass close to the axis. That's why a figure skater spins faster when they pull their arms in: they're reducing their moment of inertia without changing their angular momentum.

Moment of inertia also connects to angular momentum: $L = I \omega$.

Moment of inertia calculations

For a collection of point masses, the moment of inertia is:

$I = \sum m_i r_i^2$

Each mass $m_i$ is multiplied by the square of its distance $r_i$ from the rotation axis. This squared dependence is why mass far from the axis matters so much more.

For common solid shapes (which you'll typically be given on a formula sheet):

Notice the hoop has the largest coefficient (1) because all its mass sits at distance $R$ from the axis, while the solid sphere has the smallest (2/5) because much of its mass is closer to the center.

Parallel axis theorem: If you know the moment of inertia about an axis through the center of mass ($I_{CM}$), you can find it about any parallel axis a distance $d$ away:

$I = I_{CM} + M d^2$

For example, the thin rod's moment of inertia through one end equals $\frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2$, which matches the table above.

For composite objects (like a disk with a rod attached), calculate $I$ for each piece about the same axis, then add them together.

Rotational dynamics analysis

Here's the general approach for solving rotational dynamics problems:

1. Draw a diagram and identify the rotation axis.
2. List all forces and determine which ones create torques about that axis.
3. Calculate the net torque, assigning signs for direction.
4. Determine the total moment of inertia about the rotation axis (use the parallel axis theorem if needed for off-center components).
5. Apply $\sum \tau = I \alpha$ and solve for the unknown.
6. If the problem asks about velocities or displacements, use the rotational kinematics equations with the $\alpha$ you found.

Beyond $\sum \tau = I \alpha$, three other relationships come up frequently:

• Conservation of angular momentum: When no net external torque acts on a system, $I_1 \omega_1 = I_2 \omega_2$. If $I$ decreases, $\omega$ increases, and vice versa.
• Rotational kinetic energy: $K_r = \frac{1}{2} I \omega^2$. This is the rotational analog of $\frac{1}{2}mv^2$ and shows up in energy conservation problems involving rolling or spinning objects.
• Power in rotation: $P = \tau \omega$, analogous to $P = Fv$ in linear motion.

The work-energy theorem also applies: the net work done by torques equals the change in rotational kinetic energy, $W = \Delta K_r$.